Difference between revisions of "1992 OIM Problems/Problem 5"
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https://www.oma.org.ar/enunciados/ibe7.htm | https://www.oma.org.ar/enunciados/ibe7.htm |
Latest revision as of 08:42, 23 December 2023
Problem
The circumference and the positive numbers
and
are given so that there is a trapezoid
inscribed in
, of height
and in which the sum of the bases
and
is
. Build the trapezoid
.
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
First we realize that the trapezoid is isosceles because it is inscribed in a circle. If we draw a rhombus using the midpoints of the trapezoid, the diagonals of the rhombus will be and
We can construct such rhombus. Then, with similar triangles and the midpoints we know that the diagonal of the trapezoid is twice the length of the rhombus. Now we have all we need to start constructing:
Given the height in blue and length
in red, we begin by bisecting with straight edge and compass the line of length
and then bisecting it again as shown in the drawing above. We also bisect the blue line. Using the center of the blue line we measure
with the compass and draw a circle at the center of the point of the second bisection of the red line as shown. Then we can construct the rhombus by joining the intersection of the second perpendicular bisector with the circles and the edge of red line with the middle as shown above.
We then measure the length of the side of the rhombus with the compass and use that as a radius:
As shown above, we draw a line and using the compass with the radius of the rhombus we draw a circle with center at any point on that line so that the circle cuts the line in two places. This will be the diameter of the circle and the length of the diagonal of the trapezoid. With the compass, we measure this diagonal and use it on the given circle shown in green below:
We pick point on the circle
shown in green and using the compass with diagonal measurement we draw an arc using
as center shown in magenta, so that it cuts the green circle. This intersection is
. Using the given height as a radius on the compass we draw circles at
and
as shown. As shown above, from
we draw a line tangent to the opposite circle using straight edge. The intersection of this line with the green circle is
Likewise from
we draw a line tangent to the opposite circle. Its intersection with the green circle is
. We then draw lines connecting
and
and our trapezoid
with the conditions and the given numbers is constructed.
~Tomas Diaz. ~orders@tomasdiaz.com
- Note. I actually competed at this event in Venezuela when I was in High School representing Puerto Rico. I'm proud to say that I got full points on this one and I solved it very quickly. The only one I got full points. I had a straight rule and compass kit which I used to solve it as we're supposed to build the trapezoid with it and described the steps and how. Last week, after over 3 decades, I attempted this and spent a full 3 hours on it and couldn't solve it nor I remember what I did. Then I wrote this solution but I don't know if this was what I did back then.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.